Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict whether or not this reaction will be spontaneous at this temperature. 2NO(g) + O2(g) → 2NO2(g) ΔH = -114 kJ

Explanation:

Given: Entropy of surrounding: ΔSsurr = ?

Temperature: T= 355 K

The change in enthalpy of reaction: ΔH = -114 kJ

Pressure: P = constant

As we know, ΔH = -114 kJ ⇒ negative

Therefore, the given reaction is an exothermic reaction

Therefore, Entropy of surrounding at constant pressure is given by,

In the given reaction:

2NO(g) + O₂(g) → 2NO₂(g)

As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.

As we know, for a spontaneous process, that the total entropy should be positive.

Therefore, at the given temperature,

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

particle   p is an electron and particle q is a neutron